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A particle moves in a circle of radius 25cm covering 2 revolutions per second what will be the radial acceleration of that particle?

Answer 1

Elena Albarran

Frequency of rotation #n=2rps#

Angular velocity of the rotating particle

#omega=2pin=4pi# rad/s

Radius of the circular path

#r=25# cm.

So radial acceleration of the particle

#a_"radial"=omega^2r=(4pi)^2*25=400pi^2" "cms^-2=4pi^2~~39.48ms^-2#

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Answer 2

William Audy

#a="39.5 m/s"^2#

Formulas needed:

#C=2pir#

#a=v^2/r#

Here the circumference is calculated after everything has been converted into usable units (meters, seconds, Kelvin, etc)

#C=2pi*"0.25 m"#

#C=pi*"0.5 m"#

#C="1.571 m"#

So If the particle goes around the circle twice per second, then #(2*"1.571 m")/"1 s"# can be used to find #v#, the velocity.

If we use the equation for centripetal acceleration

#a=v^2/r#

#a = ("3.142 m/s")^2/"0.25 m" = 4xx 3.142^2 \ "m/s"^2#

#a~~"39.5 m/s"^2#

we get that the angular acceleration is equal to #39.48# meters per second per second .

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Answer 3

Olivia Askew

Radial acceleration can be calculated using the formula: ( a_r = r \cdot (\omega)^2 ), where ( r ) is the radius of the circle and ( \omega ) is the angular velocity. Plug in the values to find the radial acceleration.

( a_r = 25 , \text{cm} \cdot (2 \pi \cdot 2)^2 )

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Answer from HIX Tutor

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